Question

The solubility of ${\rm{Ca}}{{\rm{F}}_{\rm{2}}}{\rm{(}}{{\rm{K}}_{{\rm{SP}}}} = {\rm{5}}{\rm{.3 \times 1}}{{\rm{0}}^{ - 9}}{\rm{)}}$in ${\rm{0}}{\rm{.1}}\,{\rm{M}}$of ${\rm{NaF}}$solution would be :
(Assume no reaction of cation/anion).
(A) ${\rm{5}}{\rm{.3 \times 1}}{{\rm{0}}^{ - 10}}\;{\rm{M}}$
(B) ${\rm{5}}{\rm{.3 \times 1}}{{\rm{0}}^{ - 8}}\;{\rm{M}}$
(C) ${\rm{5}}{\rm{.3 \times 1}}{{\rm{0}}^{ - 7}}\;{\rm{M}}$
(D) ${\rm{5}}{\rm{.3 \times 1}}{{\rm{0}}^{ - 11}}\;{\rm{M}}$

Answer 1

As we know that, the solubility is the tendency of solute to dissolve in the solvent. Some of the compounds have greatest solubility but some of the compounds dissolve in solvent very little. So, the substances which dissolve very little are called the sparingly soluble. When the sparingly soluble salts are dissolved in water an equilibrium is established between the undissolved solid salt and ions of the dissolved salt.

Complete step by step answer
The solubility term is used to the concentration of ions in the solution. Ionic compounds have more solubility but the compounds(salts) which dissolve very less in solvent are known as sparingly soluble salts and exist in equilibrium.
The solubility product of a salt at a given temperature is equal to the product of the concentration of its ions in the saturated solution, with each concentration term raised to the power equal to the number of ions produced on dissociation of one mole of the substance. It is represented as ${{\rm{K}}_{{\rm{SP}}}}$.
If any reaction occurs as
${A_x}{B_y} \; \rightleftharpoons \; x{A}^{y+} \; + \; y{B}^{x-}$
The solubility product expression is
${{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}{\left[ {{{\rm{A}}^{{\rm{Y + }}}}} \right]^{\rm{x}}}{\left[ {{{\rm{B}}^{{\rm{x - }}}}} \right]^{\rm{y}}}$
${\rm{Ca}}{{\rm{F}}_{\rm{2}}}$is an ionic salt which is soluble in a great extent as we know from the above explanation. So, it will dissociate as
$CaF_2 \, \leftrightarrow \, Ca^{2+}(aq) \,\, + \, 2F^{-}(aq)$
If we know concentration of any ion, then we can easily calculate the solubility as-
${{\rm{K}}_{{\rm{SP}}}}{\rm{ = }}\left[ {{\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left[ {{{\rm{F}}^{\rm{ - }}}} \right]^2} - - - {\rm{(i)}}$
So, to calculate the solubility of ${\rm{Ca}}{{\rm{F}}_{\rm{2}}}$ we need the concentration of fluoride ion which will be calculated using the concentration of ${\rm{NaF}}$ as-

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\,\,\,\,\,\,\,\,{\rm{ NaF}}\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,{\rm{N}}{{\rm{a}}^{\rm{ + }}}{\rm{(aq)}} + {{\rm{F}}^{\rm{ - }}}{\rm{(aq)}}\\\ {\rm{initially}}\,\,\,\,\,\,\,\,\,\,\,\,{\rm{(0}}{\rm{.1}}\,{\rm{M)}}\,\,\,\,\;\;\;\;\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,{\rm{0}}\,\,\,\,\,\,\,\,\,\;\;\;\,\,\,\,\,\,\,\,\,{\rm{0}}\\\ {\rm{after}}\,{\rm{time(t)}}\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,{\rm{0}}{\rm{.1}}\,{\rm{M}}\,\,\,\,\,\,\;\;\,\,\,{\rm{0}}{\rm{.1}}\,{\rm{M}}\,$$

Now we have got the fluoride ion concentration as $\left[ {{{\rm{F}}^{\rm{ - }}}} \right]{\rm{ = 0}}{\rm{.1M}}$ and we have also the value of solubility product of calcium fluoride.
By putting the above values in equation ${\rm{(i)}}$, we get as-

$$\Rightarrow {\rm{5}}{\rm{.3}}\, \times {10^{ - 9}}{\rm{ = }}\left[ {{\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}} \right]{\left( {0.1} \right)^2}\\\ \Rightarrow \left[ {{\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}} \right] = \dfrac{{{\rm{5}}{\rm{.3}} \times {{10}^{ - 9}}}}{{{{(0.1)}^2}}}\\\ \Rightarrow \left[ {{\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}} \right] = {\rm{5}}{\rm{.3}} \times {10^{ - 7}}\;{\rm{M}}$$

Hence, the solubility of ${\rm{Ca}}{{\rm{F}}_{\rm{2}}}$we get as ${\rm{5}}{\rm{.3 \times 1}}{{\rm{0}}^{ - 7}}\;{\rm{M}}$.

**Therefore, the option (C) is correct.

Note:**
Knowing the solubility product of a salt, it is possible to predict whether on mixing the solution of its ions, a precipitate will be formed or not.
For the precipitation to occurs we have conditions as-
1. Ionic products are greater than ${{\rm{K}}_{{\rm{SP}}}}$, precipitation occurs.
2. Ionic product is less than ${{\rm{K}}_{{\rm{SP}}}}$, no precipitation occurs.